Chapter 2: Relational Model Tablosal (İlişkisel) Model

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1 Chapter 2: Relational Model Tablosal (İlişkisel) Model

2 Chapter 2: Relational Model
Structure of Relational Databases İlişkisel Veritabanı Yapısı Fundamental Relational-Algebra-Operations Temel İlişkisel Cebir İşlemleri Additional Relational-Algebra-Operations Ek İlişkisel Cebir İşlemleri Extended Relational-Algebra-Operations Genişletilmiş İlişkisel Cebir İşlemleri Null Values Null Değerler Modification of the Database Veritabanı Güncelleme İşlemleri

3 Tablo/İlişki (relation) örneği
tablo adı ogrenci table schema tablo şeması tablo yapısı OgrNo Adi Soyadi BolumNo 1 Ali Kurt 7 2 Ayşe Yıldız 3 Aysel Demir 6 4 Hasan Cesur 5 Ahmet Salih Zeynep Zahit kayıt/record, row/tuple/satır table istance tablo örneği tablo verisi attribute/özellik, column/sütün

4 Basic Structure Formally, given sets D1, D2, …. Dn a relation r is a subset of D1 x D2 x … x Dn Thus, a relation is a set of n-tuples (a1, a2, …, an) where each ai  Di Matematiksel olarak, bir r ilişkisi D1 x D2 x … x Dn kartezyen çarpımının bir alt kümesidir. Diğer bir ifade ile r ilişkisi aiDi olmak üzere n-değerden oluşan (a1, a2, …, an) kayıtlarının kümesidir. Example: If customer_name = {Jones, Smith, Curry, Lindsay, …} /* Set of all customer names */ customer_street = {Main, North, Park, …} /* set of all street names*/ customer_city = {Harrison, Rye, Pittsfield, …} /* set of all city names */ Then r = { (Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield) } is a relation over customer_name x customer_street x customer_city

5 Attribute Types (Veri Türleri)
Each attribute of a relation has a name. Her özellik bir isim ve tür’e sahiptir. The set of allowed values for each attribute is called the domain of the attribute Bir özelliğin alabildiği olası değerler kümesine tür/domain denir. Attribute values are (normally) required to be atomic; that is, indivisible Değerler normalde atomik yani bölünmez değerlerdir. E.g. the value of an attribute can be an account number, but cannot be a set of account numbers Domain is said to be atomic if all its members are atomic. Eğer bir tür’ün tüm elemaları atomik ise tür atomik bir türdür The special value null is a member of every domain. Null her tür’ün elemanıdır. The null value causes complications in the definition of many operations. Null birçok işlemde problem oluşturur, bu yüzden null’lı işlem durumları özel olarak tanımlanmalıdır. We shall ignore the effect of null values in our main presentation and consider their effect later. Şimdilik sunum sırasında null gözönüne alınmayacak olup ileride yeri geldiğinde null özerinde durulacaktır.

6 Relation Schema (Tablo Şeması)
A1, A2, …, An are attributes. R = (A1, A2, …, An ) is a relation schema Example: Customer_schema = (customer_name, customer_street, customer_city) r(R) denotes a relation r on the relation schema R Example: customer (Customer_schema) A1, A2, …, An özelliklerine ve D1, D2, …. Dn türlerine sahip r adlı ilişkinin şeması r(A1, A2, …, An) veya r(A1 D1, A2 D2, …, An Dn) şeklinde ifade edilir. Örneğin Ogrenci(ogrno int, adi char(10), soyadi char(10)) Ogrenci(ogrno, adi, soyadi) – veri tipleri genellikle yazılmaz

7 Relation Instance (Tablo Örneği/Verisi)
The current values (relation instance) of a relation are specified by a table. Bir ilişkinin bir tabloyla ifade edilen o anki değerlerine ilişki örneği veya ilişki verisi denir An element t of r is a tuple, represented by a row in a table. Bir r ilişkisinin t kayıdı tabloda bir satırla gösterilir. attributes (or columns) customer_name customer_street customer_city Jones Smith Curry Lindsay Main North Park Harrison Rye Pittsfield tuples (or rows) customer

8 Tabloda kayıt sırası önemli değildir.
Tablo kayıt kümesidir. Kümede eleman sırası önemsizdir. Kayıtların ilişkiye/tabloye eklenme sırası ile saklandığı dosyadaki fiziksel sırası aynı olmak zorunda değildir ve genellikle farklıdır. Buna göre, aksi söylenmediği sürece, kayıtların tablo içerisinde rastgele bir sırada tutulduğu söylenebilir. Bazen kayıtları tablo içinde sıralı tutmak istersek bunu bir SQL komutu (?) ile ayrıca ifade etmemiz gerekir. Ama kayıtları her zaman için erişirken sıralayabiliriz. Kayıtları belirli bir sırada listelemek için SELECT komutunda ORDER BY cümleciği kullanılır: SELECT * FROM ogrenci ORDER BY adi, soyadı Kayıtları sıralamanın ve tablo içinde sıralı tutmanın sıralama algoritmasından ve/veya indislemeden dolayı zaman/alan maliyeti olduğunu unutmamalıyız.

9 Database (Veritabanı)
A database consists of multiple relations. Veritabanı ilişki kümesidir. Information about an enterprise is broken up into parts, with each relation storing one part of the information. Bir konuyla ilgili bilgi kısımlara (buna veritabanı tasarımı denir) ayrılıp ayrı ilişkilerde saklanır. account : stores information about accounts depositor : stores information about which customer owns which account customer : stores information about customers Storing all information as a single relation such as bank(account_number, balance, customer_name, ..) results in repetition of information e.g.,if two customers own an account (What gets repeated?) the need for null values e.g., to represent a customer without an account Tüm bilgileri tek bir ilişkide(tabloda) saklama gereksiz veri tekrarına ve gereksiz null değerlerine sebep olur. Normalization theory (Chapter 7) deals with how to design relational schemas

10 Örnek Uygulama: Universite veritabanı
Veritabanı Şeması (Schema) ogrenci (ogrNo, adi, soyadi, bNo) – öğrenci kayıtlarını tutar hoca (hNo, adi, soyadi, bNo) – hoca kayıtlarını tutar ders (kodu, adi, kredi, bNo) – ders kayıtlarını tutar bolum (bNo, bAdi) – bolum bilgilerini tutar dersAl (ogrNo, kodu, not) – hangi öğrencinin hangi dersi aldığı bilgisini tutar dersVer (hNo, kodu) – hangi hocanın hangi dersi verdiği bilgisini tutar Veritabanı Şeması student (sid, fname, lname, did) – öğrenci kayıtlarını tutar teacher (tid, fname, lname, did) – hoca kayıtlarını tutar course (code, title, kredits, did) – ders kayıtlarını tutar department (did, dname) – bolum bilgilerini tutar take (sid, code, grade) – hangi öğrencinin hangi dersi aldığı bilgisini tutar teach (tid, code) – hangi hocanın hangi dersi verdiği bilgisini tutar

11 Universite veritabanı örneği
ogrenci ogrNo adi soyadi bNo 1 Ali Kurt 7 2 Ayşe Yıldız 3 Aysel Demir 6 hoca hno adi soyadi bno 10 Hasan Kurt 7 20 Mehmet Yıldız 30 Ozan Eren 6 ders kodu adi kredi bNo ce101 Programlama 3 7 Ce201 Veri yapısı 4 ce301 Veritabanı 2 dersAl ogrNo kodu not 2 ce101 B 1 A ce301 varlık kümesi ilişki kümesi bolum bNo bAdi 6 Bilgisayar 7 Elektrik dersVer kodu hNo ce101 10 ce301 20

12 Keys Let K  R K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R) R r ilişkisinin özellikleri kümesi ve K ise R’nin bir alt kümesi (K  R) olmak üzere, eğer K özelliklerinin değerleri olası tüm r ilişkilerinin tekil(biricik, benzersiz) bir kaydını diğerlerinden ayırt etmeye yeterli ise K r için bir süper anahtardır denir. Başka bir ifadeyle, bir ilişkideki tüm olası kayıtları diğerlerinden ayırt edebilen değerlere sahip alanlar kümesine süper anahtar denir. Ogrenci(ogrNo, adi, soyadi, dogumTarihi, dogumYeri, tcKimlik, pasaportNo) ogrNo, adi, soyadi ogrNo, adi ogrNo adi, tcKimlik tcKimlik, ogrNo adi, soyadi adi, soyadi, dogumTarihi, dogumTarihi

13 Keys (Cont.) K is a candidate key if K is minimal Eğer K süper anahtarı minimal (anahtar olmak için gereksiz alan içermiyorsa) ise, K bir aday anahtardır. ogrNo tcKimlik pasaportNo Primary key: a candidate key chosen as the principal means of identifying tuples within a relation Aday anahtarlardan seçilen bir tanesine birincil anahtar denir. Birincil anahtarlar null değeri alamaz. Should choose an attribute whose value never, or very rarely, changes. Değeri hiç değişmeyecek veya çok nadir değişen alanlar birincil anahtar seçilmelidir. E.g. address is unique, but may change. Eposta her kayıt için tekil(unique) olmakla birlikte değişebileceği için anahtar seçilmemelidir. Buna rağmen bazı uygulamalarda, uygulamanın özel ihtiyaçlarından dolayı anahtar olarak kullanılabilir.

14 Foreign Keys A relation schema may have an attribute that corresponds to the primary key of another relation. The attribute is called a foreign key. Bir ilişki başka bir ilişkinin birincil anahtarını içerebilir. Bu durumda bu alana yabancı anahtar denir. Ders-al (ogrNo, dersKodu) ilişkisinde ogrNo ve dersKodu alanları ogrenci ve ders tablolarından gelen yabancı anahtarlardır. Only values occurring in the primary key attribute of the referenced relation may occur in the foreign key attribute of the referencing relation. Bu durumda yabancı anahtar alanlarında sadece diğer tablonun birincil anahtar alanlarında geçen değerler bulunabilir. Schema diagram

15 Query Languages Language in which user requests information from the database. Kullanıcıların veritabanındaki verilere ulaşmasını sağlayan dile sorgu dili denir. Categories of languages Sorgu dili Çeşitleri Procedural İşlemin nasıl yapılacağını adım adım tarif eden diller Non-procedural, or declarative Hangi verinin istendiğinin tarif edildiği tanımsal yani adım’sal olmayan diller “Pure” languages Saf yani matematiksel diller: Relational algebra İlişkisel yani tablosal cebir Tuple relational calculus Kayıta dayalı ilişkisel matematik Domain relational calculus Tipe dayalı ilişkisel matematik Pure languages form underlying basis of query languages that people use. Matematiksel diller sorgu dillerinin altyapısını oluşturduğundan dolayı sorguların nasıl formulize edildiği, nasıl optimize edildiği ve nasıl çalıştırıldığı yönüyle önemlidir.

16 Relational Algebra Procedural language Adımsal bir dildir.
Six basic operators 6 temel işlem select:  kayıt seçme operatörü project:  sütun seçme operatörü union:  ilişki/küme/tablo bileşimi operatörü set difference: – ilişki/küme/tablo farkı operatörü Cartesian product: x katrezyen çarpım operatörü rename:  tablo ve alan isimlendirme operatörü The operators take one or two relations as inputs and produce a new relation as a result. Bu operatörler bir veya iki tabloyu giriş olarak alıp, sonuç olarak bir kayıt kümesi yani tablo döndürüler.

17 İlişkisel Cebir - SQL SELECT Bağıntısı
SELECT komutu Kayıt seçme WHERE cümleciği WHERE A=B ^ D > 5 Sütun seçme A,C (r) SELECT cümleciği SELECT A, C Kartezyen çarpım r x s FROM cümleciği FROM r, s Birleşim r  s UNION cümleciği r UNION s Fark r – s EXCEPT/MINUS cümleciği r EXCEPT s Tablo ve alan isimlendirme p x(a1, a2, ... an) (r) FROM/SELECT cümleciği (AS) SELECT b1 AS a1, ... Bn AS an FROM r AS x A=B ^ D > 5 (r)

18 İlişkisel Cebir- SQL Bağıntısı
İlişkisel cebir ve SQL kabaca aynı işi yapar. İkiside sorgu dilidir. Birbirlerine ifade gücü bakımından kabaca eşdeğer sayılabilirler. İlişkisel cebir ve SQL ifadeleri birbirlerine çevrilebilir. SQL komutları çalıştırılırken işlemleri optimize etmek için ve çalıştırmak için ilişkisel cebir kullanılır. İlişkisel cebirde ifadelerin sonucu daima bir ilişkidir; bundan dolayı sonuçtaki tekrarlı kayıtlar bir defa listelenir. Buna duplicate elimination denir. SQLde ise genellikle sorguların sonucundaki tekrarlı kayıtların hepsi listelenir. Tekrarlı kayıtları bir defa listelemek için SELECT cümleciğinde DISTINCT anahtar kelimesi kullanılır. DISTINCT kullanıldığında kayıtlar sıralanacağı için sonuçların daha geç geleceği bilinmelidir.

19 Select Operation – Example
Relation r A B C D     1 5 12 23 7 3 10 Örne sorgu: r ilişkisinde A ve B alanlarında aynı değere ve D alanında 5’ten büyük değerler içeren kayıtları listeleyiniz. A=B ^ D > 5 (r) A B C D     1 23 7 10

20 Select Operation Notation:  p(r) p is called the selection predicate
Defined as: p(r) = {t | t  r and p(t)} Where p is a formula in propositional calculus consisting of terms connected by :  (and),  (or),  (not) Each term is one of: <attribute> op <attribute> or <constant> where op is one of: =, , >, . <.  Example of selection:  branch_name=“Perryridge”(account)

21 Project Operation – Example
B C Relation r:   10 20 30 40 1 2 A,C (r) A C A C   1 2   1 2 =

22 Project Operation Notation:
where A1, A2 are attribute names and r is a relation name. The result is defined as the relation of k columns obtained by erasing the columns that are not listed Duplicate rows removed from result, since relations are sets Example: To eliminate the branch_name attribute of account account_number, balance (account)

23 Union Operation – Example
Relations r, s: A B A B   1 2   2 3 s r A B   1 2 3 r  s:

24 Union Operation Notation: r  s Defined as:
r  s = {t | t  r or t  s} For r  s to be valid. 1. r, s must have the same arity (same number of attributes) 2. The attribute domains must be compatible (example: 2nd column of r deals with the same type of values as does the 2nd column of s) Example: to find all customers with either an account or a loan customer_name (depositor)  customer_name (borrower)

25 Set Difference Operation – Example
Relations r, s: A B A B   1 2   2 3 s r r – s: A B   1

26 Set Difference Operation
Notation r – s Defined as: r – s = {t | t  r and t  s} Set differences must be taken between compatible relations. r and s must have the same arity attribute domains of r and s must be compatible

27 Cartesian-Product Operation – Example
Relations r, s: A B C D E   1 2    10 20 a b r s r x s: A B C D E   1 2    10 20 a b

28 Cartesian-Product Operation
Notation r x s Defined as: r x s = {t q | t  r and q  s} Assume that attributes of r(R) and s(S) are disjoint. (That is, R  S = ). If attributes of r(R) and s(S) are not disjoint, then renaming must be used.

29 Composition of Operations
Can build expressions using multiple operations Example: A=C(r x s) r x s A=C(r x s) A B C D E   1 2      10 20 a b A B C D E   1 2   10 20 a b

30 Rename Operation Allows us to name, and therefore to refer to, the results of relational-algebra expressions. Allows us to refer to a relation by more than one name. Example:  x (E) returns the expression E under the name X If a relational-algebra expression E has arity n, then returns the result of expression E under the name X, and with the attributes renamed to A1 , A2 , …., An .

31 Alan seçme İşlemi Örnek sorgu: Üniversite veritabanındaki öğrencilerin soyadlarını listeleyiniz. fname, lname (student) – tekil değerler(kayıtlar ir defa listeleinir SELECT fname, lname FROM student Örnek sorgu:

32 Banking Example branch (branch_name, branch_city, assets)
customer (customer_name, customer_street, customer_city) account (account_number, branch_name, balance) loan (loan_number, branch_name, amount) depositor (customer_name, account_number) borrower (customer_name, loan_number)

33 Example Queries amount > 1200 (loan)
Find all loans of over $1200 amount > 1200 (loan) Find the loan number for each loan of an amount greater than $1200 loan_number (amount > 1200 (loan)) Find the names of all customers who have a loan, an account, or both, from the bank customer_name (borrower)  customer_name (depositor)

34 Example Queries customer_name (branch_name=“Perryridge”
Find the names of all customers who have a loan at the Perryridge branch. customer_name (branch_name=“Perryridge” (borrower.loan_number = loan.loan_number(borrower x loan))) Find the names of all customers who have a loan at the Perryridge branch but do not have an account at any branch of the bank. customer_name (branch_name = “Perryridge” (borrower.loan_number = loan.loan_number(borrower x loan))) – customer_name(depositor)

35 Example Queries Find the names of all customers who have a loan at the Perryridge branch. Query 1 customer_name (branch_name = “Perryridge” ( borrower.loan_number = loan.loan_number (borrower x loan))) Query 2 customer_name(loan.loan_number = borrower.loan_number ( (branch_name = “Perryridge” (loan)) x borrower))

36 Example Queries balance(account) - account.balance
Find the largest account balance Strategy: Find those balances that are not the largest Rename account relation as d so that we can compare each account balance with all others Use set difference to find those account balances that were not found in the earlier step. The query is: balance(account) - account.balance (account.balance < d.balance (account x rd (account)))

37 Formal Definition  x (E1), x is the new name for the result of E1
A basic expression in the relational algebra consists of either one of the following: A relation in the database A constant relation Let E1 and E2 be relational-algebra expressions; the following are all relational-algebra expressions: E1  E2 E1 – E2 E1 x E2 p (E1), P is a predicate on attributes in E1 s(E1), S is a list consisting of some of the attributes in E1  x (E1), x is the new name for the result of E1

38 Additional Operations (Ek İşlemler)
We define additional operations that do not add any power to the relational algebra, but that simplify common queries. Be 4 ek işlem ilişkisel cebirin ifade gücünü arttırmasa da çoğu sorguları basitlerştirdikleri için tanımlanmıştır. Set intersection İlişki/küme/tablo kesişimi Natural join Doğal kartezyen/biRleşim (ortak alan ve değerle kartezyen çarpım) Division İlişki/Küme/tablo Bölümü Assignment Atama

39 Set-Intersection Operation
Notation: r  s Defined as: r  s = { t | t  r and t  s } Assume: r, s have the same arity attributes of r and s are compatible Note: r  s = r – (r – s)

40 Set-Intersection Operation – Example
Relation r, s: r  s A B A B   1 2   2 3 r s A B 

41 Natural-Join Operation
Notation: r s Let r and s be relations on schemas R and S respectively. Then, r s is a relation on schema R  S obtained as follows: Consider each pair of tuples tr from r and ts from s. If tr and ts have the same value on each of the attributes in R  S, add a tuple t to the result, where t has the same value as tr on r t has the same value as ts on s Example: R = (A, B, C, D) S = (E, B, D) Result schema = (A, B, C, D, E) r s is defined as: r.A, r.B, r.C, r.D, s.E (r.B = s.B  r.D = s.D (r x s))

42 Natural Join Operation – Example
Relations r, s: B D E A B C D 1 3 2 a b          1 2 4    a b r s r s A B C D E   1 2    a b   

43 Division Operation r  s Notation:
Suited to queries that include the phrase “for all”. Let r and s be relations on schemas R and S respectively where R = (A1, …, Am , B1, …, Bn ) S = (B1, …, Bn) The result of r  s is a relation on schema R – S = (A1, …, Am) r  s = { t | t   R-S (r)   u  s ( tu  r ) } Where tu means the concatenation of tuples t and u to produce a single tuple

44 Division Operation – Example
Relations r, s: A B B      1 2 3 4 6 1 2 s r  s: A r  

45 Another Division Example
Relations r, s: A B C D E D E    a    a b 1 3 a b 1 s r r  s: A B C   a 

46 Division Operation (Cont.)
Property Let q = r  s Then q is the largest relation satisfying q x s  r Definition in terms of the basic algebra operation Let r(R) and s(S) be relations, and let S  R r  s = R-S (r ) – R-S ( ( R-S (r ) x s ) – R-S,S(r )) To see why R-S,S (r) simply reorders attributes of r R-S (R-S (r ) x s ) – R-S,S(r) ) gives those tuples t in R-S (r ) such that for some tuple u  s, tu  r.

47 Assignment Operation The assignment operation () provides a convenient way to express complex queries. Write query as a sequential program consisting of a series of assignments followed by an expression whose value is displayed as a result of the query. Assignment must always be made to a temporary relation variable. Example: Write r  s as temp1  R-S (r ) temp2  R-S ((temp1 x s ) – R-S,S (r )) result = temp1 – temp2 The result to the right of the  is assigned to the relation variable on the left of the . May use variable in subsequent expressions.

48 Bank Example Queries Find the names of all customers who have a loan and an account at bank. customer_name (borrower)  customer_name (depositor) Find the name of all customers who have a loan at the bank and the loan amount customer_name, loan_number, amount (borrower loan)

49 Bank Example Queries Find all customers who have an account from at least the “Downtown” and the Uptown” branches. Query 1 customer_name (branch_name = “Downtown” (depositor account ))  customer_name (branch_name = “Uptown” (depositor account)) Query 2 customer_name, branch_name (depositor account)  temp(branch_name) ({(“Downtown” ), (“Uptown” )}) Note that Query 2 uses a constant relation.

50 Bank Example Queries Find all customers who have an account at all branches located in Brooklyn city. customer_name, branch_name (depositor account)  branch_name (branch_city = “Brooklyn” (branch))

51 Extended Relational-Algebra-Operations Genişletilmiş İlişkisel Cebir İşlemleri
Generalized Projection Genelleşmiş sütun seçme Aggregate Functions Grup Fonksiyonları Outer Join Dış doğal kartezyen (dış birleşim)

52 Generalized Projection
Extends the projection operation by allowing arithmetic functions to be used in the projection list. E is any relational-algebra expression Each of F1, F2, …, Fn are are arithmetic expressions involving constants and attributes in the schema of E. Given relation credit_info(customer_name, limit, credit_balance), find how much more each person can spend: customer_name, limit – credit_balance (credit_info)

53 Aggregate Functions and Operations
Aggregation function takes a collection of values and returns a single value as a result. avg: average value min: minimum value max: maximum value sum: sum of values count: number of values Aggregate operation in relational algebra E is any relational-algebra expression G1, G2 …, Gn is a list of attributes on which to group (can be empty) Each Fi is an aggregate function Each Ai is an attribute name

54 Aggregate Operation – Example
Relation r: A B C     7 3 10 g sum(c) (r) sum(c ) 27

55 Aggregate Operation – Example
Relation account grouped by branch-name: branch_name account_number balance Perryridge Brighton Redwood A-102 A-201 A-217 A-215 A-222 400 900 750 700 branch_name g sum(balance) (account) branch_name sum(balance) Perryridge Brighton Redwood 1300 1500 700

56 Aggregate Functions (Cont.)
Result of aggregation does not have a name Can use rename operation to give it a name For convenience, we permit renaming as part of aggregate operation branch_name g sum(balance) as sum_balance (account)

57 Outer Join null signifies that the value is unknown or does not exist
An extension of the join operation that avoids loss of information. Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result of the join. Uses null values: null signifies that the value is unknown or does not exist All comparisons involving null are (roughly speaking) false by definition. We shall study precise meaning of comparisons with nulls later

58 Outer Join – Example Relation loan 3000 4000 1700 loan_number amount
branch_name Downtown Redwood Perryridge Relation borrower customer_name loan_number Jones Smith Hayes L-170 L-230 L-155

59 Outer Join – Example Join loan borrower loan_number amount L-170 L-230
3000 4000 customer_name Jones Smith branch_name Downtown Redwood Left Outer Join loan borrower Jones Smith null loan_number amount L-170 L-230 L-260 3000 4000 1700 customer_name branch_name Downtown Redwood Perryridge

60 Outer Join – Example Right Outer Join loan borrower loan_number amount
3000 4000 null customer_name Jones Smith Hayes branch_name Downtown Redwood Full Outer Join loan borrower loan_number amount L-170 L-230 L-260 L-155 3000 4000 1700 null customer_name Jones Smith Hayes branch_name Downtown Redwood Perryridge

61 Null Values It is possible for tuples to have a null value, denoted by null, for some of their attributes null signifies an unknown value or that a value does not exist. The result of any arithmetic expression involving null is null. Aggregate functions simply ignore null values (as in SQL) For duplicate elimination and grouping, null is treated like any other value, and two nulls are assumed to be the same (as in SQL)

62 Null Values Comparisons with null values return the special truth value: unknown If false was used instead of unknown, then not (A < 5) would not be equivalent to A >= 5 Three-valued logic using the truth value unknown: OR: (unknown or true) = true, (unknown or false) = unknown (unknown or unknown) = unknown AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown NOT: (not unknown) = unknown In SQL “P is unknown” evaluates to true if predicate P evaluates to unknown Result of select predicate is treated as false if it evaluates to unknown

63 Modification of the Database
The content of the database may be modified using the following operations: Deletion Insertion Updating All these operations are expressed using the assignment operator.

64 Deletion A delete request is expressed similarly to a query, except instead of displaying tuples to the user, the selected tuples are removed from the database. Can delete only whole tuples; cannot delete values on only particular attributes A deletion is expressed in relational algebra by: r  r – E where r is a relation and E is a relational algebra query.

65 Deletion Examples Delete all account records in the Perryridge branch. account  account – branch_name = “Perryridge” (account ) Delete all loan records with amount in the range of 0 to 50 loan  loan – amount 0and amount  50 (loan) Delete all accounts at branches located in Needham. r1  branch_city = “Needham” (account branch ) r2   account_number, branch_name, balance (r1) r3   customer_name, account_number (r2 depositor) account  account – r2 depositor  depositor – r3

66 Insertion To insert data into a relation, we either:
specify a tuple to be inserted write a query whose result is a set of tuples to be inserted in relational algebra, an insertion is expressed by: r  r  E where r is a relation and E is a relational algebra expression. The insertion of a single tuple is expressed by letting E be a constant relation containing one tuple.

67 Insertion Examples account  account  {(“A-973”, “Perryridge”, 1200)}
Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account  account  {(“A-973”, “Perryridge”, 1200)} depositor  depositor  {(“Smith”, “A-973”)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. r1  (branch_name = “Perryridge” (borrower loan)) account  account  loan_number, branch_name, 200 (r1) depositor  depositor  customer_name, loan_number (r1)

68 Updating A mechanism to change a value in a tuple without charging all values in the tuple Use the generalized projection operator to do this task Each Fi is either the I th attribute of r, if the I th attribute is not updated, or, if the attribute is to be updated Fi is an expression, involving only constants and the attributes of r, which gives the new value for the attribute

69 Update Examples Make interest payments by increasing all balances by 5 percent. account   account_number, branch_name, balance * 1.05 (account) Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent account   account_number, branch_name, balance * 1.06 ( BAL  (account ))   account_number, branch_name, balance * 1.05 (BAL  (account))

70 End of Chapter 2

71 Figure 2.3. The branch relation

72 Figure 2.6: The loan relation

73 Figure 2.7: The borrower relation

74 Figure 2.9 Result of branch_name = “Perryridge” (loan)

75 Figure 2.10: Loan number and the amount of the loan

76 Figure 2.11: Names of all customers who have either an account or an loan

77 Figure 2.12: Customers with an account but no loan

78 Figure 2.13: Result of borrower |X| loan

79 Figure 2.14

80 Figure 2.15

81 Figure 2.16

82 Figure 2.17 Largest account balance in the bank

83 Figure 2.18: Customers who live on the same street and in the same city as Smith

84 Figure 2.19: Customers with both an account and a loan at the bank

85 Figure 2.20

86 Figure 2.21

87 Figure 2.22

88 Figure 2.23

89 Figure 2.24: The credit_info relation

90 Figure 2.25

91 Figure 2.26: The pt_works relation

92 Figure 2.27 The pt_works relation after regrouping

93 Figure 2.28

94 Figure 2.29

95 Figure 2.30 The employee and ft_works relations

96 Figure 2.31

97 Figure 2.32

98 Figure 2.33

99 Figure 2.34